I am looking at a metric which is defined as (Eq 2.4 Glampedakis & Babak)

$$ g_{\mu \nu} = g_{\mu \nu}^K + \epsilon h_{\mu \nu}$$

where $g_{\mu \nu}^K$ is the original unperturbed metric (Kerr) and $h_{\mu \nu}$ some perturbation.

Now, I know that the indices of the perturbation are raised/lowered using the unperturbed metric i.e.

$$ h_{\alpha \beta} = g_{\alpha \gamma}^K g_{\beta \delta}^K h^{\gamma \delta}$$

(see e.g. application in Eq 2. of Narzilloev et al. 2019)

My question is how to get the contravariant form of $g^{\mu \nu}$?

Option 1 is that simply,

$$ g^{\mu \nu} = g^{\mu \nu}_K + \epsilon h^{\mu \nu}$$

Option 2 considers $g^{\mu \nu}$ as an independent matrix and we invert it the usual way.

However these two options seem to not be equivalent. For example consider the $g^{03}$ term.

Option 1 tells us that $$ g^{03} = g^{0 3}_K + \epsilon h^{0 3}$$ but $h^{0 3} = 0$ and so $g^{03} = g^{0 3}_K$.

But Option 2 tells us that

$$ g^{03} = - \frac{g_{03}}{ \tilde{g}} = \frac{-1}{\tilde{g}} (g_{03}^{K} + \epsilon h_{03}) \ne g^{0 3}_K$$

where $\tilde{g} = g_{00} g_{33} - g_{03}^2$ and we have exploited the symmetries of the matrix (e.g. Eq 19.13 of these notes)

Can anyone provide some guidance on where I am going wrong? Thanks.

  • $\begingroup$ To first order in $\epsilon$, $ g^{\mu \nu} = g^{\mu \nu}_K - \epsilon h^{\mu \nu}$. $\endgroup$
    – G. Smith
    Sep 3, 2019 at 16:32
  • $\begingroup$ Can this be shown by inverting the covariant perturbed metric? $\endgroup$ Sep 3, 2019 at 16:34
  • $\begingroup$ It can be shown by computing $g_{\mu\lambda}g^{\lambda\nu}$. The $O(\epsilon)$ terms cancel, leaving $\delta_\mu^\nu+O(\epsilon^2)$. $\endgroup$
    – G. Smith
    Sep 3, 2019 at 16:40
  • $\begingroup$ I've answered a similar question here - physics.stackexchange.com/a/330277/133418. Also, note that using the unperturbed metric to raise/lower indices is just because we're using perturbation theory and we neglect higher order terms. In principle, and in full generality, one should use the full metric. $\endgroup$
    – Avantgarde
    Sep 4, 2019 at 14:53

1 Answer 1


Oooh you're in bad luck. This is quite possibly one of the worst aspect of the perturbation formalism in general relativity, and quite possibly the reason the full non-linear theory isn't commonly used.

There is no finite expression for the inverse metric of $g$. This is due to the binomial inverse theorem. What we want is to find the inverse of the matrix $g + h$. From this theorem, this is

$$(g + h)^{-1} = g^{-1} - g^{-1} (I + hg^{-1})^{-1} h g^{-1}$$

From the form of the expansion, you can immediatly see the issue : the definition is recursive. Any expansion will require further expansion. There are many forms you can expand it into (here's another one), but in the end it is always an infinite sum.

As we usually deal with this formalism to linearize the theory, though, it is usually cut off at the linear term. This will give us

$$(g + h)^{-1} = g^{-1} - g^{-1} h g^{-1} = g^{\mu\nu} - h^{\mu\nu}$$


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